# Image and Kernel check please

• August 28th 2010, 09:06 AM
Image and Kernel check please
T(x1,x2,x3,x4) = (4x1-x2, 5x2 + 3x4, -x3)

Image = (4,0,0) (-1,5,0) (0,0,-1) (0,3,0)

So the vectors that form a basis are (4,0,0) (0,0,-1) (0,3,0)

Kernel

4x1 - x2 = 0
5x2 + 3x4 = 0
-x3 = 0

x1 = 1/4 x2
x2 = x2
x3 = 0
x4 = -5/3 x2

(x1, x2, x3, x4) = (1/4x2, x2, 0, -5/3 x2)

= x2(1/4, 1, 0, -5/3)

Kernel = (1/4, 1, 0, -5/3)
• August 29th 2010, 10:39 AM
(Actually, any two of (4, 0, 0), (0, 3, 0), and (-1, 5, 0), together with (0, 0, -1) would be a basis. In fact, just (1, 0, 0), (0, 1, 0), (0, 0, 1) is a basis because the image is all of $R^3$.)